Track / Stage 6 / 85 of 400 #1085 of 1964

Problem 1085

National olympiad, first round

Geometry Difficulty 6.1 Prove it

Question 2 As shown in Figure 3, $O$ is the circumcenter of acute $\triangle ABC$ , $AH$ is the altitude, point $P$ lies on line $AO$ , and $PD$ , $PE$ , $PF$ are the angle bisectors of $\angle BPC$ , $\angle CPA$ , $\angle APB$ respectively. Prove that points $D$ , $E$ , $F$ , $H$ are concyclic.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

This article only proves the case where point $P$ is inside $\triangle A B C$ . As for the case where point $P$ is outside the triangle, readers can refer to the following proof to complete the argument.

In fact, it is not easy to directly prove that points $D, E, F, H$ are concyclic.

Therefore, we redefine point $H$ . Let the circumcircle $\Gamma_{1}$ of $\triangle D E F$ intersect $B C$ at another point $H$ . We only need to prove: $A H \perp B C$ .
Let the circle $\Gamma_{1}$ intersect $A C$ and $A B$ at points $S$ and $T$ , respectively.
Notice,
$\begin{array}{l} \frac{B D \cdot B H}{C D \cdot C H} \cdot \frac{C E \cdot C S}{A E \cdot A S} \cdot \frac{A F \cdot A T}{B F \cdot B T}=1, \\ \frac{B D}{C D} \cdot \frac{C E}{A E} \cdot \frac{A F}{B F}=\frac{P B}{P C} \cdot \frac{P C}{P A} \cdot \frac{P A}{P B}=1 . \\ \text { Hence } \frac{B H}{H C} \cdot \frac{C S}{S A} \cdot \frac{A T}{T B}=1 . \end{array}$

By Ceva's Theorem, we know that $B S, C T, A H$ are concurrent, and we denote this point as $W$ .
At this point, let's pause and look at a classic problem.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.